Understanding Deutch's AlgorithmHow would I implement the quantum oracle in Deutsch's algorithm?Deutsch Algorithm on a Quantum Turing MachineHow is the Deutsch-Jozsa algorithm faster than classical for practical implementation?Deutsch algorithm with equal input bitsWhy doesn't Deutsch-Jozsa Algorithm show that P ≠ BQP?How to understand Deutsch-Jozsa algorithm from an adiabatic perspective?Deutsch–Jozsa algorithm: why is $f$ constant?How does an oracle function in Grover's algorithm actually work?What can be a mini research project based on Grover's algorithm or the Deutsch Jozsa algorithm?Deutsch-Jozsa algorithm as a generalization of Bernstein-Vazirani

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Understanding Deutch's Algorithm


How would I implement the quantum oracle in Deutsch's algorithm?Deutsch Algorithm on a Quantum Turing MachineHow is the Deutsch-Jozsa algorithm faster than classical for practical implementation?Deutsch algorithm with equal input bitsWhy doesn't Deutsch-Jozsa Algorithm show that P ≠ BQP?How to understand Deutsch-Jozsa algorithm from an adiabatic perspective?Deutsch–Jozsa algorithm: why is $f$ constant?How does an oracle function in Grover's algorithm actually work?What can be a mini research project based on Grover's algorithm or the Deutsch Jozsa algorithm?Deutsch-Jozsa algorithm as a generalization of Bernstein-Vazirani






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








1












$begingroup$


I am reading John Waltrous notes from his course CPSC 519 on quantum computing. In a pre-discussion before presenting Deutsch's algorithm to determine whether a function is constant or not, the author presents a function $B_f |x rangle |y rangle = |x rangle |y oplus f(x) rangle $, and the diagram:
enter image description here



The inital state is $|0 rangle |1 rangle$, and after the first two Hadamard transforms, will be $big(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1ranglebig)big(frac 1 sqrt 2 |0rangle-frac 1 sqrt 2 |1ranglebig) $.



Up to this far I understand. The author then writes: "After performing the $B_f$ operation the state is transformed to :



$frac 1 2 |0 rangle big(|0 oplus f(0)rangle - |1 oplus f(0)ranglebig) + frac 1 2 |1 rangle big(|0 oplus f(1)rangle) - |1 oplus f(1) ranglebig)$ ".



I am not sure how this was obtained, from what I understand, the operation should be
$frac 1 sqrt 2 big( |0rangle + |1ranglebig) otimes big|(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1rangle) oplus f(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1rangle) bigrangle$ (simply subbing in $x,y$ to $B_f$). Any insights appreciated as this subject is completely new to me, although I have a decent mathematics and computer science background.










share|improve this question









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    $begingroup$


    I am reading John Waltrous notes from his course CPSC 519 on quantum computing. In a pre-discussion before presenting Deutsch's algorithm to determine whether a function is constant or not, the author presents a function $B_f |x rangle |y rangle = |x rangle |y oplus f(x) rangle $, and the diagram:
    enter image description here



    The inital state is $|0 rangle |1 rangle$, and after the first two Hadamard transforms, will be $big(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1ranglebig)big(frac 1 sqrt 2 |0rangle-frac 1 sqrt 2 |1ranglebig) $.



    Up to this far I understand. The author then writes: "After performing the $B_f$ operation the state is transformed to :



    $frac 1 2 |0 rangle big(|0 oplus f(0)rangle - |1 oplus f(0)ranglebig) + frac 1 2 |1 rangle big(|0 oplus f(1)rangle) - |1 oplus f(1) ranglebig)$ ".



    I am not sure how this was obtained, from what I understand, the operation should be
    $frac 1 sqrt 2 big( |0rangle + |1ranglebig) otimes big|(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1rangle) oplus f(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1rangle) bigrangle$ (simply subbing in $x,y$ to $B_f$). Any insights appreciated as this subject is completely new to me, although I have a decent mathematics and computer science background.










    share|improve this question









    New contributor



    IntegrateThis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$














      1












      1








      1





      $begingroup$


      I am reading John Waltrous notes from his course CPSC 519 on quantum computing. In a pre-discussion before presenting Deutsch's algorithm to determine whether a function is constant or not, the author presents a function $B_f |x rangle |y rangle = |x rangle |y oplus f(x) rangle $, and the diagram:
      enter image description here



      The inital state is $|0 rangle |1 rangle$, and after the first two Hadamard transforms, will be $big(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1ranglebig)big(frac 1 sqrt 2 |0rangle-frac 1 sqrt 2 |1ranglebig) $.



      Up to this far I understand. The author then writes: "After performing the $B_f$ operation the state is transformed to :



      $frac 1 2 |0 rangle big(|0 oplus f(0)rangle - |1 oplus f(0)ranglebig) + frac 1 2 |1 rangle big(|0 oplus f(1)rangle) - |1 oplus f(1) ranglebig)$ ".



      I am not sure how this was obtained, from what I understand, the operation should be
      $frac 1 sqrt 2 big( |0rangle + |1ranglebig) otimes big|(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1rangle) oplus f(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1rangle) bigrangle$ (simply subbing in $x,y$ to $B_f$). Any insights appreciated as this subject is completely new to me, although I have a decent mathematics and computer science background.










      share|improve this question









      New contributor



      IntegrateThis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$




      I am reading John Waltrous notes from his course CPSC 519 on quantum computing. In a pre-discussion before presenting Deutsch's algorithm to determine whether a function is constant or not, the author presents a function $B_f |x rangle |y rangle = |x rangle |y oplus f(x) rangle $, and the diagram:
      enter image description here



      The inital state is $|0 rangle |1 rangle$, and after the first two Hadamard transforms, will be $big(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1ranglebig)big(frac 1 sqrt 2 |0rangle-frac 1 sqrt 2 |1ranglebig) $.



      Up to this far I understand. The author then writes: "After performing the $B_f$ operation the state is transformed to :



      $frac 1 2 |0 rangle big(|0 oplus f(0)rangle - |1 oplus f(0)ranglebig) + frac 1 2 |1 rangle big(|0 oplus f(1)rangle) - |1 oplus f(1) ranglebig)$ ".



      I am not sure how this was obtained, from what I understand, the operation should be
      $frac 1 sqrt 2 big( |0rangle + |1ranglebig) otimes big|(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1rangle) oplus f(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1rangle) bigrangle$ (simply subbing in $x,y$ to $B_f$). Any insights appreciated as this subject is completely new to me, although I have a decent mathematics and computer science background.







      deutsch-jozsa-algorithm






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      IntegrateThis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|improve this question




      share|improve this question








      edited 3 hours ago









      Mariia Mykhailova

      2,0751212




      2,0751212






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      asked 3 hours ago









      IntegrateThisIntegrateThis

      1084




      1084




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          1 Answer
          1






          active

          oldest

          votes


















          4












          $begingroup$

          Remember that when you define the oracle effect as $B_f |x rangle |y rangle = |x rangle |y oplus f(x) rangle $, $f(x)$ is a classical function of a classical 1-bit argument, so you do not have a way to compute $f(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1rangle)$ (a function of a quantum state).



          The quantum oracles that implement classical functions are defined as follows:



          1. Define the effect of the oracle on all basis states for $|xrangle$ and $|yrangle$: $B_f |x rangle |y rangle = |x rangle |y oplus f(x) rangle $.


          2. This will automatically define the effect of the oracle on all superposition states: the oracle is a quantum operation and has to be linear in the state on which it acts. So if you start with a state $frac12 (|00rangle + |10rangle - |01rangle - |11rangle)$ (which is the state after applying Hadamard gates) and apply the oracle, you need to apply oracle to each basis state separately. You'll get


          $$B_f frac12 (|00rangle + |10rangle - |01rangle - |11rangle) = frac12 (B_f|00rangle + B_f|10rangle - B_f|01rangle - B_f|11rangle) =$$



          $$ = frac12 (|0rangle|0 oplus f(0)rangle + |1rangle|0 oplus f(1)rangle - |0rangle|1 oplus f(0)rangle - |1rangle|1 oplus f(1)rangle)$$



          Which is the same as the expression in the notes, up to a different grouping or terms.




          The part about the oracles being defined by their effect on basis states is implicit in a lot of sources I've seen, and is a frequent source of confusion. If you need more mathematical details on this, we ended up writing it up here.






          share|improve this answer









          $endgroup$








          • 2




            $begingroup$
            Another resource that may be helpful is Learn Quantum Computing with Python and Q# which should have the chapter on Deutsch–Jozsa algorithm up shortly! We work though implementing Deutsch–Jozsa in Q# as well as in Python with QuTiP. <manning.com/books/…>
            $endgroup$
            – Dr. Sarah Kaiser
            3 hours ago











          • $begingroup$
            Thanks so much for your help! I am very grateful :)
            $endgroup$
            – IntegrateThis
            2 hours ago











          Your Answer








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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes









          4












          $begingroup$

          Remember that when you define the oracle effect as $B_f |x rangle |y rangle = |x rangle |y oplus f(x) rangle $, $f(x)$ is a classical function of a classical 1-bit argument, so you do not have a way to compute $f(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1rangle)$ (a function of a quantum state).



          The quantum oracles that implement classical functions are defined as follows:



          1. Define the effect of the oracle on all basis states for $|xrangle$ and $|yrangle$: $B_f |x rangle |y rangle = |x rangle |y oplus f(x) rangle $.


          2. This will automatically define the effect of the oracle on all superposition states: the oracle is a quantum operation and has to be linear in the state on which it acts. So if you start with a state $frac12 (|00rangle + |10rangle - |01rangle - |11rangle)$ (which is the state after applying Hadamard gates) and apply the oracle, you need to apply oracle to each basis state separately. You'll get


          $$B_f frac12 (|00rangle + |10rangle - |01rangle - |11rangle) = frac12 (B_f|00rangle + B_f|10rangle - B_f|01rangle - B_f|11rangle) =$$



          $$ = frac12 (|0rangle|0 oplus f(0)rangle + |1rangle|0 oplus f(1)rangle - |0rangle|1 oplus f(0)rangle - |1rangle|1 oplus f(1)rangle)$$



          Which is the same as the expression in the notes, up to a different grouping or terms.




          The part about the oracles being defined by their effect on basis states is implicit in a lot of sources I've seen, and is a frequent source of confusion. If you need more mathematical details on this, we ended up writing it up here.






          share|improve this answer









          $endgroup$








          • 2




            $begingroup$
            Another resource that may be helpful is Learn Quantum Computing with Python and Q# which should have the chapter on Deutsch–Jozsa algorithm up shortly! We work though implementing Deutsch–Jozsa in Q# as well as in Python with QuTiP. <manning.com/books/…>
            $endgroup$
            – Dr. Sarah Kaiser
            3 hours ago











          • $begingroup$
            Thanks so much for your help! I am very grateful :)
            $endgroup$
            – IntegrateThis
            2 hours ago















          4












          $begingroup$

          Remember that when you define the oracle effect as $B_f |x rangle |y rangle = |x rangle |y oplus f(x) rangle $, $f(x)$ is a classical function of a classical 1-bit argument, so you do not have a way to compute $f(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1rangle)$ (a function of a quantum state).



          The quantum oracles that implement classical functions are defined as follows:



          1. Define the effect of the oracle on all basis states for $|xrangle$ and $|yrangle$: $B_f |x rangle |y rangle = |x rangle |y oplus f(x) rangle $.


          2. This will automatically define the effect of the oracle on all superposition states: the oracle is a quantum operation and has to be linear in the state on which it acts. So if you start with a state $frac12 (|00rangle + |10rangle - |01rangle - |11rangle)$ (which is the state after applying Hadamard gates) and apply the oracle, you need to apply oracle to each basis state separately. You'll get


          $$B_f frac12 (|00rangle + |10rangle - |01rangle - |11rangle) = frac12 (B_f|00rangle + B_f|10rangle - B_f|01rangle - B_f|11rangle) =$$



          $$ = frac12 (|0rangle|0 oplus f(0)rangle + |1rangle|0 oplus f(1)rangle - |0rangle|1 oplus f(0)rangle - |1rangle|1 oplus f(1)rangle)$$



          Which is the same as the expression in the notes, up to a different grouping or terms.




          The part about the oracles being defined by their effect on basis states is implicit in a lot of sources I've seen, and is a frequent source of confusion. If you need more mathematical details on this, we ended up writing it up here.






          share|improve this answer









          $endgroup$








          • 2




            $begingroup$
            Another resource that may be helpful is Learn Quantum Computing with Python and Q# which should have the chapter on Deutsch–Jozsa algorithm up shortly! We work though implementing Deutsch–Jozsa in Q# as well as in Python with QuTiP. <manning.com/books/…>
            $endgroup$
            – Dr. Sarah Kaiser
            3 hours ago











          • $begingroup$
            Thanks so much for your help! I am very grateful :)
            $endgroup$
            – IntegrateThis
            2 hours ago













          4












          4








          4





          $begingroup$

          Remember that when you define the oracle effect as $B_f |x rangle |y rangle = |x rangle |y oplus f(x) rangle $, $f(x)$ is a classical function of a classical 1-bit argument, so you do not have a way to compute $f(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1rangle)$ (a function of a quantum state).



          The quantum oracles that implement classical functions are defined as follows:



          1. Define the effect of the oracle on all basis states for $|xrangle$ and $|yrangle$: $B_f |x rangle |y rangle = |x rangle |y oplus f(x) rangle $.


          2. This will automatically define the effect of the oracle on all superposition states: the oracle is a quantum operation and has to be linear in the state on which it acts. So if you start with a state $frac12 (|00rangle + |10rangle - |01rangle - |11rangle)$ (which is the state after applying Hadamard gates) and apply the oracle, you need to apply oracle to each basis state separately. You'll get


          $$B_f frac12 (|00rangle + |10rangle - |01rangle - |11rangle) = frac12 (B_f|00rangle + B_f|10rangle - B_f|01rangle - B_f|11rangle) =$$



          $$ = frac12 (|0rangle|0 oplus f(0)rangle + |1rangle|0 oplus f(1)rangle - |0rangle|1 oplus f(0)rangle - |1rangle|1 oplus f(1)rangle)$$



          Which is the same as the expression in the notes, up to a different grouping or terms.




          The part about the oracles being defined by their effect on basis states is implicit in a lot of sources I've seen, and is a frequent source of confusion. If you need more mathematical details on this, we ended up writing it up here.






          share|improve this answer









          $endgroup$



          Remember that when you define the oracle effect as $B_f |x rangle |y rangle = |x rangle |y oplus f(x) rangle $, $f(x)$ is a classical function of a classical 1-bit argument, so you do not have a way to compute $f(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1rangle)$ (a function of a quantum state).



          The quantum oracles that implement classical functions are defined as follows:



          1. Define the effect of the oracle on all basis states for $|xrangle$ and $|yrangle$: $B_f |x rangle |y rangle = |x rangle |y oplus f(x) rangle $.


          2. This will automatically define the effect of the oracle on all superposition states: the oracle is a quantum operation and has to be linear in the state on which it acts. So if you start with a state $frac12 (|00rangle + |10rangle - |01rangle - |11rangle)$ (which is the state after applying Hadamard gates) and apply the oracle, you need to apply oracle to each basis state separately. You'll get


          $$B_f frac12 (|00rangle + |10rangle - |01rangle - |11rangle) = frac12 (B_f|00rangle + B_f|10rangle - B_f|01rangle - B_f|11rangle) =$$



          $$ = frac12 (|0rangle|0 oplus f(0)rangle + |1rangle|0 oplus f(1)rangle - |0rangle|1 oplus f(0)rangle - |1rangle|1 oplus f(1)rangle)$$



          Which is the same as the expression in the notes, up to a different grouping or terms.




          The part about the oracles being defined by their effect on basis states is implicit in a lot of sources I've seen, and is a frequent source of confusion. If you need more mathematical details on this, we ended up writing it up here.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 3 hours ago









          Mariia MykhailovaMariia Mykhailova

          2,0751212




          2,0751212







          • 2




            $begingroup$
            Another resource that may be helpful is Learn Quantum Computing with Python and Q# which should have the chapter on Deutsch–Jozsa algorithm up shortly! We work though implementing Deutsch–Jozsa in Q# as well as in Python with QuTiP. <manning.com/books/…>
            $endgroup$
            – Dr. Sarah Kaiser
            3 hours ago











          • $begingroup$
            Thanks so much for your help! I am very grateful :)
            $endgroup$
            – IntegrateThis
            2 hours ago












          • 2




            $begingroup$
            Another resource that may be helpful is Learn Quantum Computing with Python and Q# which should have the chapter on Deutsch–Jozsa algorithm up shortly! We work though implementing Deutsch–Jozsa in Q# as well as in Python with QuTiP. <manning.com/books/…>
            $endgroup$
            – Dr. Sarah Kaiser
            3 hours ago











          • $begingroup$
            Thanks so much for your help! I am very grateful :)
            $endgroup$
            – IntegrateThis
            2 hours ago







          2




          2




          $begingroup$
          Another resource that may be helpful is Learn Quantum Computing with Python and Q# which should have the chapter on Deutsch–Jozsa algorithm up shortly! We work though implementing Deutsch–Jozsa in Q# as well as in Python with QuTiP. <manning.com/books/…>
          $endgroup$
          – Dr. Sarah Kaiser
          3 hours ago





          $begingroup$
          Another resource that may be helpful is Learn Quantum Computing with Python and Q# which should have the chapter on Deutsch–Jozsa algorithm up shortly! We work though implementing Deutsch–Jozsa in Q# as well as in Python with QuTiP. <manning.com/books/…>
          $endgroup$
          – Dr. Sarah Kaiser
          3 hours ago













          $begingroup$
          Thanks so much for your help! I am very grateful :)
          $endgroup$
          – IntegrateThis
          2 hours ago




          $begingroup$
          Thanks so much for your help! I am very grateful :)
          $endgroup$
          – IntegrateThis
          2 hours ago










          IntegrateThis is a new contributor. Be nice, and check out our Code of Conduct.









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          Smell Mother Skizze Discussion Tachometer Jar Alligator Star 끌다 자세 의문 과학적t Barbaric The round system critiques the connection. Definition: A wind instrument of music in use among the Spaniards Nasty Level 이상 분노 금년 월급 근교 Cloth Owner Permissible Shock Purring Parched Raise 오전 장면 햄 서투르다 The smash instructs the squeamish instrument. Large Nosy Nalpure Chalk Travel Crayon Bite your tongue The Hulk 신호 대사 사과하다 The work boosts the knowledgeable size. Steeplump Level Wooden Shake Teaching Jump 이제 복도 접다 공중전화 부지런하다 Rub Average Ruthless Busyglide Glost oven Didelphia Control A fly on the wall Jaws 지하철 거